Clipper problem

Thread Starter

phuzionz

Joined Dec 5, 2008
47
Dear members,

When you have a clipper circuit with two diodes in serie to the power supply. I try to draw it. The two diodes are with the cathode to the Vs+

Vs +
|
|
|
diode
|
|
|
|
diode
|
|
|
GND

This circuit will clip the voltage, depend on what is the power supply value. for instance, a Vs from 5V, then the upper diode wil clip at 5.7V and the lower at -5.7.
But how can the lower diode clips the negative voltage to -5.7. The anode is alaways 0V, thus you must conduct at -0.7V.
Can somebody explain me that

Thank you,
Jef
 

Ron H

Joined Apr 14, 2005
7,063
Dear members,

When you have a clipper circuit with two diodes in serie to the power supply. I try to draw it. The two diodes are with the cathode to the Vs+

Vs +
|
|
|
diode
|
|
|
|
diode
|
|
|
GND

This circuit will clip the voltage, depend on what is the power supply value. for instance, a Vs from 5V, then the upper diode wil clip at 5.7V and the lower at -5.7.
But how can the lower diode clips the negative voltage to -5.7. The anode is alaways 0V, thus you must conduct at -0.7V.
Can somebody explain me that

Thank you,
Jef
It doesn't work as you described. Furthermore, if the power supply is not current limited, the diodes will be destroyed. I think you drew the circuit wrong. Where did you get this information? Can you post a corrected graphic schematic in .GIF or .PNG format?
 

DrNick

Joined Dec 13, 2006
110
You are correct. It will clip at 5.7V and -0.7V, not 5.7 and -5.7. Now if you changed that ground to a -5V supply you could achieve -5.7V clipping.

Ron H, I think he is intending on connecting the circuit as follws:

+5V
|
|
Cathode
|
Anode
|
|
signal to be clamped------>
|
|
Cathode
|
Anode
|
|
GND
 

Ron H

Joined Apr 14, 2005
7,063
You are correct. It will clip at 5.7V and -0.7V, not 5.7 and -5.7. Now if you changed that ground to a -5V supply you could achieve -5.7V clipping.

Ron H, I think he is intending on connecting the circuit as follws:

+5V
|
|
Cathode
|
Anode
|
|
signal to be clamped------>
|
|
Cathode
|
Anode
|
|
GND
That makes more sense.
 

Thread Starter

phuzionz

Joined Dec 5, 2008
47
Nick you're right, sorry for the bad picture. The lower diode is there to protect the µC input to negative voltages. Can you really destroy the input with a negative voltage?

But in most circuits to protect a µC you see this circuit added with a zener. What is the prupose of that zener then?

Jef
 

eblc1388

Joined Nov 28, 2008
1,542
Can you really destroy the input with a negative voltage?
Yes.

But in most circuits to protect a µC you see this circuit added with a zener. What is the prupose of that zener then?

Jef
The same two diodes in your first post already exists on the pins of many microcontrollers inside the chip. However, it has been mentioned that user should not "deliberately" design the input circuit to pass any current through these diodes. A maximum current rating is not provided in the datasheet and passing any current is considered "out of Spec" to the part.

Using a series resistor with a zener diode to ground externally would provide the same protection as two diodes. For high positive voltage, the zener limits the input voltage to the zener voltage by conducting and clips the negative voltage to 0.7V below ground for negative voltage acting like a normal diode.
 

Ron H

Joined Apr 14, 2005
7,063
Yes.



The same two diodes in your first post already exists on the pins of many microcontrollers inside the chip. However, it has been mentioned that user should not "deliberately" design the input circuit to pass any current through these diodes. A maximum current rating is not provided in the datasheet and passing any current is considered "out of Spec" to the part.
This is not true for PICs. Electrical Specifications of the 16F628A (I believe all PICs have these same highlighted specs - see attached) have what I believe to be conflicting specifications. 20mA through an input diode would certainly have a voltage drop greater than 0.3V. I suspect it means that one should not connect a voltage source directly to an input pin if that source exceeds either rail by more than 0.3V, but a source limited to <20mA is OK. Microchip published app note AN521 which specifically recommends using the input diodes as clamps for applications such as power line zero crossing detection.
 

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Thread Starter

phuzionz

Joined Dec 5, 2008
47
Using a series resistor with a zener diode to ground externally would provide the same protection as two diodes. For high positive voltage, the zener limits the input voltage to the zener voltage by conducting and clips the negative voltage to 0.7V below ground for negative voltage acting like a normal diode.
So what is then the advantage to use zenerdiode with series resistor instead of two diodes to Vs.

But i thought that it was also very important to limit the current through the diodes. How can you do that, because 20mA is nit so much?
 

Ron H

Joined Apr 14, 2005
7,063
So what is then the advantage to use zenerdiode with series resistor instead of two diodes to Vs.

But i thought that it was also very important to limit the current through the diodes. How can you do that, because 20mA is nit so much?
A series resistor is a great current limiter.:)
 

DrNick

Joined Dec 13, 2006
110
If you are trying to protect a uC, I would recommend that you use an active clamp, consisting of op-amps and diodes.

These circuits can perform relatively precision clamping to any desired voltage. The disadvantage is that you quadruple the number of components needed. If you can't find anything on active clamping on the web, just ask me to draw up a little circuit for you. Too lazy at the moment hehe...
 

Ron H

Joined Apr 14, 2005
7,063
If you are trying to protect a uC, I would recommend that you use an active clamp, consisting of op-amps and diodes.

These circuits can perform relatively precision clamping to any desired voltage. The disadvantage is that you quadruple the number of components needed. If you can't find anything on active clamping on the web, just ask me to draw up a little circuit for you. Too lazy at the moment hehe...
Good for an analog signals such as A/D inputs. Maybe not so good for some digital signals...:)
 

eblc1388

Joined Nov 28, 2008
1,542
I never recommended high currents. All I can do is once again reference AN521. Perhaps Microchip's right hand doesn't know what the left hand is doing.:D
That AN521 is dated 1997 and this one 2008. Go figure.:confused:

However, on the AVR microcontroller chips, no such current rating is given in the datasheet and it is often mentioned that it would be bad practice to try to pass current through these ESD diodes. Always use an external clamp.
 

Ron H

Joined Apr 14, 2005
7,063
That AN521 is dated 1997 and this one 2008. Go figure.:confused:

However, on the AVR microcontroller chips, no such current rating is given in the datasheet and it is often mentioned that it would be bad practice to try to pass current through these ESD diodes. Always use an external clamp.
You would think Microchip would remove AN521 from their web site. Maybe I'll try to contact the app engineers and see if they have a comment. Can you post a link to the excerpt you posted (always a good practice)?
 
Last edited:

Ron H

Joined Apr 14, 2005
7,063
However, on the AVR microcontroller chips, no such current rating is given in the datasheet and it is often mentioned that it would be bad practice to try to pass current through these ESD diodes. Always use an external clamp.
Have a look at AVR182. It recommends the same technique as Microchip's AN521.:confused:
 

eblc1388

Joined Nov 28, 2008
1,542
Do let us know what is the outcome of this enquiry.

But a side question, the clamp(series R+ESD diodes) only works if there is load drawing some current on the VCC line. Now these uC can all go to SLEEP and consumes less a uA or less current drawn from the VCC bus.

Would forming a clamp with a high series resistor have the possibilities of raising the VCC potential over the allowable limit of the chip?
 

Thread Starter

phuzionz

Joined Dec 5, 2008
47
If you are trying to protect a uC, I would recommend that you use an active clamp, consisting of op-amps and diodes.

These circuits can perform relatively precision clamping to any desired voltage. The disadvantage is that you quadruple the number of components needed. If you can't find anything on active clamping on the web, just ask me to draw up a little circuit for you. Too lazy at the moment hehe...
I have found a circuitry

http://www.eetasia.com/ARTICLES/2007AUG/C/EEOL_2007AUG14_POW_EMD_NP_01.JPG

Jef
 
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